高等数学复旦大学出版社习题答案一
习题一
1. 下列函数是否相等, 为什么
?
(1)f (x ) =(3)f (x ) =
g (x ) =x ; (2)y =sin (3x +1), u =sin (3t +1); x -1x -1
22
2
, g (x ) =x +1.
解: (1)相等.
因为两函数的定义域相同, 都是实数集R ;
由两函数相等.
(2)相等.
因为两函数的定义域相同, 都是实数集R , 由已知函数关系式显然可得两函数的对应法则也相同, 所以两函数相等.
(3)不相等.
因为函数f (x ) 的定义域是{x x ∈R , x ≠1}, 而函数g (x ) 的定义域是实数集R , 两函数的定义域不同, 所以两函数不相等. 2. 求下列函数的定义域
(1)y =(3)y =
arctan x x -1
2
=x 知两函数的对应法则也相同; 所以
1x
; (2)y =1lg(1-x )
;
; (4) y =arccos(2sin x ).
解: (1)要使函数有意义, 必须
⎧4-x ≥0⎧x ≤4
即 ⎨ ⎨
x ≠0x ≠0⎩⎩
所以函数的定义域是(-∞, 0) (0,4].
(2)要使函数有意义, 必须
⎧x +3≥0
⎪
⎨lg(1-x ) ≠0 即 ⎪1-x >0⎩
⎧x ≥-3⎪
⎨x ≠0 ⎪x
所以函数的定义域是[-3,0)∪(0,1).
(3)要使函数有意义, 必须
2
x -1≠0 即 x ≠±1
所以函数的定义域是(-∞, -1) (-1,1) (1,+∞) .
(4)要使函数有意义, 必须
-1≤2sin x ≤1 即 -
12
≤sin x ≤
12
即-
π6
+2k π≤x ≤π6
+k π≤x ≤
π6π6
+2k π或
5π6
+2k π≤x ≤
7π6
+2k π,(k 为整数).
也即-+k π (k 为整数). π6+k π,
π6
+k π], k为整数.
所以函数的定义域是[-
1⎧sin , ⎪
3. 求函数y =⎨x
⎪0, ⎩
x ≠0x =0
的定义域与值域.
解: 由已知显然有函数的定义域为(-∞, +∞), 又当x ≠0时, 时, sin
1x
1x
可以是不为零的任意实数, 此
可以取遍[-1,1]上所有的值, 所以函数的值域为[-1,1].
1-x 1+x
4. 没f (x ) =, 求f (0),f (-x ), f ().
x
1
解: f (0)=
1-01+0
=1, f (-x ) =
1-(-x ) 1+(-x )
=
1+x
1, f () =
x 1-x
=x -1.
1x +11+
x
1-
1
5. 设f (x ) =⎨
⎧1, ⎩x +1,
-1≤x
, 求f (x -1) .
⎧1,
解: f (x -1) =⎨
⎩(x -1) +1,
x
-1≤x -1
1, 0≤x
=⎨.
x , 1≤x ≤3⎩
6. 设f (x ) =2, g (x ) =x ln x , 求f (g (x )), g (f (x )), f (f (x )) 和g (g (x )) . 解: f (g (x )) =2
g (x )
=2
x ln x
,
x
x
x
g (f (x )) =f (x ) ln f (x ) =2⋅ln 2=(x ln 2) ⋅2,
f (f (x )) =2
f (x )
=2,
2
x
g (g (x )) =g (x ) ln g (x ) =x ln x ln(x ln x ).
3
7. 证明:f (x ) =2x -
1和g (x ) =
.
3
证:由y =2x -
1解得x =
故函数f (x ) =2x 3-
1的反函数是y =
x ∈R ) ,
这与g (x ) =
是同一个函
数, 所以f (x ) =2x 3-
1和g (x ) =
.
8. 求下列函数的反函数及其定义域:
(1)y =
1-x 1+x
2x +5
; (2)y =ln(x +2) +1; ; (4)y =1+cos x , x ∈[0,π].
3
(3)y =3
解: (1)由y =
1-x 1+x
解得x =
1-y 1+y
,
1-x 1+x
所以函数y =
1-x 1+x
的反函数为y =(x ≠-1) .
(2)由y =ln(x +2) +1得x =e y -1-2,
所以, 函数y =ln(x +2) +1的反函数为y =e x -1-2 (x ∈R ) .
(3)由y =32x +5解得x =
12
(log3y -5)
12
(log3x -5) (x >0) .
所以, 函数y =32x +5的反函数为y =
(4)由y =1+
cos 3x 得cos x =
, 又x ∈[0,π],
故x =arccos .
3
又由-1≤cos x ≤1得0≤1+cos x ≤2,
即0≤y ≤2, 故可得反函数的定义域为[0,2],所以, 函数y =1+cos x , x ∈[0,π]的反函
数为y =arccos
3
(0≤x ≤2) .
9. 判断下列函数在定义域内的有界性及单调性:
(1)y =
x 1+x
2
; (2)y =x +ln x
x 1+x
2
解: (1)函数的定义域为(-∞,+∞), 当x ≤0时, 有故∀x ∈(-∞, +∞), 有y ≤又因为函数y =
x 1+x
2
≤0, 当x >0时, 有
x 1+x
2
≤
x 2x
=
12
,
12
. 即函数y =
x 1+x
2
有上界.
为奇函数, 所以函数的图形关于原点对称, 由对称性及函数有上界知, 函
x 1+x
2
数必有下界, 因而函数y =有界.
又由y 1-y 2=
x 11+x
21
-
x 21+x
22
=
(x 1-x 2)(1-x 1x 2) (1+x )(1+x )
21
22
知, 当x 1>x 2且x 1x 2y 2, 而
当x 1>x 2且x 1x 2>1时, y 1
x 1+x
2
在定义域内不单调.
(2)函数的定义域为(0,+∞),
∀M >0, ∃x 1>0且x 1>M ; ∃x 2>e
M
>0, 使ln x 2>M .
取x 0=max{x 1, x 2}, 则有x 0+ln x 0>x 1+ln x 2>2M >M , 所以函数y =x +ln x 在定义域内是无界的. 又当0
故y 1-y 2=(x 1+ln x 1) -(x 2+ln x 2) =(x 1-x 2) +(lnx 1-ln x 2)
(1)f (x ) =
(2)y =e
2x
-e
-2x
+sin x .
解
: (1)
f (-x ) =
∴f (x ) =
=
+
-2x
=f (x )
.
2x
(2) f (-x ) =e
∴函数y =e
-e +sin(-x ) =e
-2x
-e
2x
+sin x =-(e
2x
-e
-2x
+sin x ) =-f (x )
2x
-e
-2x
+sin x 是奇函数.
11. 设f (x ) 定义在(-∞,+∞) 上, 证明:
(1) f (x ) +f (-x ) 为偶函数; (2)f (x ) -f (-x ) 为奇函数. 证: (1)设F (x ) =f (x ) +f (-x ) , 则∀x ∈(-∞, +∞) , 有F (-x ) =f (-x ) +f (x ) =F (x ) 故f (x ) +f (-x ) 为偶函数.
(2)设G (x ) =f (x ) -f (-x ), 则∀x ∈(-∞, +∞) ,
有G (-x ) =f (-x ) -f (-x ) =-[f (x ) -f (-x )]=-G (x )
故f (x ) -f (-x ) 为奇函数.
12. 某厂生产某种产品, 年销售量为106件, 每批生产需要准备费103元, 而每件的年库存费为0.05元, 如果销售是均匀的, 求准备费与库存费之和的总费用与年销售批数之间的函数(销售均匀是指商品库存数为批量的一半). 解: 设年销售批数为x , 则准备费为10x ;
又每批有产品
10x
6
3
件, 库存数为
10
6
2x
件, 库存费为
10
6
2x
⨯0.05元.
设总费用为, 则y =10x +
3
10⨯0.05
2x
6
.
13. 邮局规定国内的平信, 每20g 付邮资0.80元, 不足20 g按20 g计算, 信件重量不得超过2kg, 试确定邮资y 与重量x 的关系. 解: 当x 能被20整除, 即[
x 20]=
x 20x 20
时, 邮资y =
x 20
x 20
⨯0.80=
x 25
;
当x 不能被20整除时, 即[]≠时, 由题意知邮资y =
⎡x ⎤
⨯0.80. +1⎢⎥⎣20⎦
⎧x
⎪25, ⎪
综上所述有y =⎨
⎪⎡x +1⎤⨯0.80, ⎢⎥⎪⎦⎩⎣20
0
x ⎡x ⎤
=; ⎢⎥20⎣20⎦x ⎡x ⎤
≠. ⎢⎣20⎥⎦20
其中
x x ⎡x ⎤⎡x ⎤
+1的最大整数. , 分别表示不超过, +1⎢⎥⎢⎥2020⎣20⎦⎣20⎦
14. 已知水渠的横断面为等腰梯形, 斜角ϕ=40°, 如图所示. 当过水断面ABCD 的面积为定值S 0时, 求湿周L (L =AB +BC +CD ) 与水深h 之间的函数关系式, 并指明其定义域
.
图1-1
解:
S 0=
12
h (A D +B C ) =
12
h (2h cot ϕ+B C +B C ) =h (B C +h cot ϕ)
从而 B C =
S 0h
-h cot ϕ.
L =AB +BC +C D (AB =C D ) =2
h sin ϕ+
+BC =2
h sin ϕS 0h
+S 0h
-h cot ϕ
=
S 0h
2-cos ϕsin ϕ
h =+
2-cos 40sin 40
h
由h >0, B C =
S 0h
-h cot ϕ>
0得定义域为.
15. 下列函数是由哪些基本初等函数复合而成的?
1
(1)y =(1+x 2) 4; (2)y =sin 2
(1+2x );
1
(3)y =(1+10
-x
5
) 2; (4)y =
1 1+arcsin 2x
.
1
1
解: (1)y =(1+x 2) 4是由y =u 4, u =1+x 2复合而成.
(2)y =sin 2(1+2x ) 是由y =u 2, u =sin v , v =1+2x 复合而成.
1
1
(3)y =(1+10-x
5
) 2是由y =u 2, u =1+v , v =10w , w =-x 5
复合而成.
(4)y =111+arcsin 2x
是由y =u -, u =1+v , v =arcsin w , w =2x 复合而成.
16. 证明:
(1)arcsin h x =ln(x + (2)arctan h x =
11+x 2
ln
1-x
, -1
证: (1)由y =sinh x =
e x
-e 2
得e 2x -2y e x -1=0
解方程e 2x -2y e x -1=
0得e x =y ±因为e x >0,
所以e x =y +
x =ln(y +
所以y =
sinh x 的反函数是y =arcsin h x =ln(x + (-∞
-x (2)由y =tanh x =
e -e e x +e
-x
得e 2x =
1+y 1+y 1-y
, 得2x =ln
1+y 1-y
, x =
12
ln
1-y
;
又由
1+y 1-y
>0得-1
所以函数y =tanh x 的反函数为
y =arctan h x =
1x 2ln 1+1-x
-1
17. 写出下列数列的通项公式, 并观察其变化趋势:
(1) 0,
13, 24, 35, 46, ; (2) 1, 0, -3, 0, 5, 0, -7, 0, ; (3) -3, 579
3, -5, 7, . 解: (1)x n -1n =
n +1
当n →∞时, x n →1. (2)x n -1n =n cos
2
π,
当n 无限增大时, 有三种变化趋势:趋向于+∞, 趋向于0, 趋向于-∞.
(3)x n =(-1)
n
2n +12n -1
, 当n 无限增大时, 变化趁势有两种, 分别趋于1,-1.
18. 对下列数列求a =lim x n , 并对给定的ε确定正整数N (ε) , 使对所有n >N (ε) , 有
n →∞
x n -a
ε:
(1)x n =
1n sin
n π2
, ε=0.001; (2)x n =
ε=0.0001.
解: (1)a =lim x n =0, ∀ε>0, 要使x n -0=
n →∞
1n
sin
n π2
1n
1
ε
. 取N =
⎡1⎤
, 则⎢⎥⎣ε⎦
当n >N 时, 必有x n -0
⎡1⎤
=1000或大于1000的整数.
⎢⎥⎣0.001⎦
(2)a =lim x n =0, ∀ε>0,
要使x n -0=
n →∞
=
2
2=
1
只要>
1
ε
即n >
1
ε
2
即可.
取N =
⎡1⎤
, 则当n >N 时, 有x n -0
1⎡⎤88
或大于10的整数. =102⎥⎢⎣0.0001⎦
当ε=0.0001时, N =
19. 根据数列极限的定义证明:
(1)lim
1n
2
n →∞
=0; (2)lim n
3n -12n +1
n →∞
=
32
;
(3)lim
n →∞
n 个
=1; (4)lim 0.99 9=1.
n →∞
证: (1) ∀ε>0, 要使
1n
2
-0=
1n
2
只要n >
.
取N =, 则当n>N时, 恒有1n
2
-0
1n
2
n →∞
=0.
(2) ∀ε>0, 要使
3n -12n +1
-
32
=
52(2n +1)
54n
5n
5
ε
, 取N =
⎡5⎤
, 则当⎢⎣ε⎥⎦
n>N时, 恒有
3n -12n +1
-
32
3n -12n +1
n →∞
=
32
.
(3) ∀ε>0, 要
使
1=
2
a n
22
要n >
,
取
n =, 则当n>N时,
1
从而lim n →∞
n
=1.
(4)因为对于所有的正整数n , 有
n 个 0, 不防设ε
n 个1-ln ε⎡-ln ε⎤ 只要取=, N =, 则当n >N
⎢⎥0. 99 9-110n ln 10⎣ln 10⎦
时, 恒有
n 个
n 个
20. 若lim x n =a , 证明lim x n =a , 并举反例说明反之不一定成立.
n →∞
n →∞
证: lim x n =0, 由极限的定义知, ∀ε>0, ∃N >0, 当n >N 时, 恒有x n -a
n →∞
而 x n -a
∴∀ε>0, ∃N >0, 当n >N 时, 恒有x n -a
由极限的定义知lim x n =a .
n →∞
但这个结论的逆不成立. 如x n =(-1) , lim x n =1, 但lim x n 不存在.
n →∞
n →∞
n
21. 利用单调有界准则证明下列数列有极限, 并求其极限值
:
(1)x 1=
x n +1=
n =1, 2, ; (2)x 1=1, x n +1=1+
x n 1+x n
, n =1, 2, .
证
: (1) x 1=
x k +1=
=2.
故对所有正整数n 有x n
又x n +1-x n =x n =
从而x n +1-x n >0即x n +1>x n ,
>0, 又由x n
2即数列{x n }是单调递增的.
由极限的单调有界准则知, 数列{x n }有极限. 设lim x n =a ,
则a =
, 于是n →∞
a 2=2a , a =2, a =0(不合题意, 舍去), ∴lim x n =2.
n →∞
(2) 因为x 1=1>0, 且x n +1=1+
x n 1+x ,
n
所以
, 即数列有界
又 x ⎛
+x n
⎫⎛
x n -1
⎫
x n -x
n -1
n +1-x n = 1⎝1+x ⎪- 1+n ⎭⎝
1+x ⎪=
n -1⎭(1+x n )(1+x n -1) 由1+x n >0,1+x n -1>0知x n +1-x n 与x n -x n -1同号， 从而可推得x n +1-x n 与x 2-x 1同号， 而 x 131=1, x 2=1+2
=
2
, x 2-x >1
0 故x n +1-x n >0, 即x n +1>x n
所以数列{x n }单调递增，由单调有界准则知，{x n }的极限存在. 设lim x n =a , 则a =1+
a
n →∞
1+a
，解得
a =
1+1-2
, a =2
(不合题意，舍去).
所以
lim 1+n →∞
x n =
2
22. 用函数极限定义证明：
2
2
(1)lim
sin x 3x -1x →+∞x
=0; (2)lim
x →∞
x 2
+4
=3; (3)lim x -4x →-2
x +2
=-4;
2
(4)lim
1-4x
1
x +1
=2; (5)lim x →0
x sin
x
=0.
x →-
12
2证：(1)∀ε>0, 要使
sin x sin x 1x
-0=
x
≤
x 只须x >
1
ε
, 取X >
1
ε
，则当x >X 时，必有
sin x x
-0
故lim
sin x x
x →+∞
=0.
(2)∀ε>0, 要使
3x -1x +4
22
-3=
13x +4
2
13|x |
2
只须x >
取X =
x >X 时，必有
3x 2
-1,
x 2+4
-3
故lim
3x -1x →∞
x 2
+4
=3.
(3) ∀ε>0, 要使
x 2
-4x +2
-(-4) =x +2
只要取δ=ε，则 当0
x 2
-4x +2
-(-4)
2
故x lim
x -4→-2
x +2
=-4.
(4) ∀ε>0, 要使
1-4x
2
ε,
2x +1
-2=2x +1=2x +
12
12
ε
2
，取δ=
ε
2
，则
当0
14x
2
2
1-2x +1
-2
2
故lim
1-4x
x →-
12
2x +1
=2.
(5) ∀ε>0, 要使
x sin
1x
-0=x sin
1x
≤x
只要取δ=ε，则
当0
1x
-0
故lim x sin
1x →0
x
=0.
23. 求下列极限：
2
(1)lim
x -32
x x 2+1
x →3
; (2)lim
x +x →1
x 4
-3x 2
+1; 2(3)lim
x -1x 3-x
2x 2-x -1 x →∞
; (4)lim
x →∞
x 4
-3x 2
+1
;
(5)lim
x 2
+12x +1
+1)(n +2)(n +3)
x →∞
; (6)lim
(n n →∞
5n
3
;
(7)若lim ⎛ x 2+1⎫1
2x +1-ax -b ⎪=，求a 和b .
x →∞
⎝⎭2
lim 解：(1)lim
x 2
-3x →3
(x 2
+1
=
x 2-3)
=
9-3x →3
lim x →3
(x 2+1)9+1
=
35
.
2lim 2
2
x →1
(x +x ) (2)lim
x +x x →1
x 4
-3x 2
+1
=
lim (x 4
2
=
1+1x →1
-3x +1)
14
-3⨯12
+1
=-2.
121-
(3)lim
x -1x 2
x →∞
2x 2
-x -1
=lim
x →∞
=
12-
1-12
.
x x 2
1
3
-
1
⎛1⎫3lim (4)lim x -x x x x →∞ 1⎝x - x 3⎪
⎭x →∞x 4-3x 2
+1=lim x →∞1-31=⎛31⎫
=0. x 2+x 4lim x →∞ ⎝
1-x 2+x 4⎪
⎭2
+
1lim ⎛(5) lim
2x +1x
2
2
+1⎫x →∞
x 2
+1
=lim x
x →∞
=x →∞
⎝x x 2⎪⎭1+
1x
2
lim ⎛
x →∞ ⎝
1+1⎫=0
x 2⎪⎭由无穷大与无穷小的关系知， l x 2
+1
x →i ∞2x +1=∞. (6)lim
(n +1)(n +2)(n +3)
n 5n
3
5lim ⎛
n →∞ ⎝1+1⎫n ⎪⎛2⎫⎛3⎫→∞
=
1
⎭ ⎝1+n ⎪⎭ ⎝1+n ⎪⎭
=1
5lim ⎛n →∞ ⎝1+1⎫⎛2⎫⎛
3⎫1n ⎪⎭⋅lim n →∞ ⎝1+n ⎪⎭⋅lim n →∞ ⎝1+n ⎪⎭=5
.
24. 解:因为
x +1x +1
2
-ax -b =
(1-a ) x -(a +b ) x +(1-b )
x +1
2
⎛x 2+1⎫1
由已知lim -ax -b ⎪=知，分式的分子与分母的次数相同，且x 项的系数之
x →∞
⎝x +1⎭2
比为
12
，于是
1-a =0 且
-(a +b )
11
=
2
解得 a =1, b =-
32
.
25. 利用夹逼定理求下列数列的极限:
(1)lim[(n +1) k
n →∞
-n k
],0
1;
(2)lim
n →∞
其中a 1, a 1, , a m 为给定的正常数
;
1
(3)lim n →∞
(1+2n
+3n
) n ;
(4)lim
n →∞
解: (1) 0
1
⎢
⎣
) k
-1⎤k ⎡1⎤1⎥
而lim 0=0, 当k
1=0
n →∞
n →∞
n
1-k
∴lim[(k k
n →∞
n +1) -n ]=0.
(2)记a =m ax{a 1, a 2, , a m } 则有
1
即
a
1
而 l i m a =a , m
n ⋅l a i =m a n →∞
n →∞
,
故
lim
n →∞
=a
即
l i m a a x {n →∞
=
1a , 2
a , m , . }
11
1
(3) (3n
) n
+3n
) n
) n
1n +1n
即 3
n +1
n n
3
=而 l i m 3
n →∞
3,
n
n →∞n
l i n m =3
1
n
3=) .
3
故 l i m (+1
n →∞
2+
(4) 1
1n
0,
1l i +m 1= ) n →∞n
1
而 l i m =1
n →∞
故
l i n →∞
=. 1
26. 通过恒等变形求下列极限：
(1)lim
1+2+3+ +(n -1)
n
2
2
n →∞
11⎛
; (2)lim 1++
+n
n →∞
⎝22
x -6x +8x -5x +4
2
3
⎫
⎪; ⎭
(3)lim
x -2x +1x -1
3
2
x →1
2
; (4)lim
x →4
2
;
(5)lim x 2
x →+∞
-; (6)lim x →0
(7)lim
x →5
lim
x →
1-cot x 2-cot x -cot x (1-
2
3
π4
;
(9)lim (1+x )(1+x ) (1+x ) (x
(10)lim
x →∞
22
n
-
(1-
n -1
x →1
(1-x )
;
3⎛1
(11)lim -3x →1
⎝1-x 1-x (13)lim
log a (1+x )
x
3
x -x +1⎫
; (12)lim ; ⎪2
x →1(x -1) ⎭
a -1x x
x
x →0
; (14)lim
x →0
; .
(15)lim (1+2x ) sin x ; (16)lim ln
x →0
x →0
sin x
解：(1)lim
1+2+3+ +(n -1)
n
2
n →∞
=lim
n (n -1) 2n
2
n →∞
=lim
n →∞
1⎛1⎫1 1-⎪=. 2⎝n ⎭2
⎛1⎫1- ⎪
11⎫⎛⎝2⎭
(2)lim 1++ +n ⎪=lim
n →∞1⎝22⎭n →∞
1-
2(3)lim
x -2x +1x -1x -6x +8x -5x +4
3222
x →1
n +1
=2.
=lim
(x -1) x -1
2
x →1
=lim (x -1) =0.
x →1
=23.
(4)lim
x →4
=lim
(x -2)(x -4) (x -1)(x -4)
x →4
=lim
x -2x -1
x →4
(5)lim x 2
x →+∞
2
-
=x lim →+∞
=lim
4
x →+∞
=2.
(6)lim
x →0
=lim
x →0
-x
=-lim (1+=-2.
x →0
(7)lim
x →5
=lim x →5
=lim
x →5
=
=
=lim
3x →5(8)lim
x →
1-cot x 2-cot x -cot x
3
3
π4
=lim
x →
1-cot x
(1-cot x ) +(1-cot x ) (1-cot x )(1+cot x +cot x ) (1-cot x )(1+1+cot x +cot x ) 1+cot x +cot x 2+cot x +cot x
22
22
3
π4
=lim
x →
π4
=lim
x →
π4
n
=
34
.
(9)lim (1+x )(1+x ) (1+x ) (x
x →∞
22
=lim =lim
(1-x )(1+x )(1+x ) (1+x )
1-x
1-x
2
n +1
22
n
x →∞
x →∞
1-x
=
11-x
.
(10)lim
(1--
(1-
n -1
n -1x →1
(1-x )
=lim
x →1
=lim
x →1
=
1
12⨯3⨯4⨯ ⨯n
=
n ! .
22
(11)lim ⎛1+x +x -3x +x -x →1 1
⎝1-x -3⎫1-x 3⎪⎭=lim 2x →1(1-x )(1+x +x 2) =lim x →1(1-x )(1+x +x 2)
=lim
(x -1)(x +2) x →1
(1-x )(1+x +x 2
)
=lim
-(x +2) x →1
1+x +x
2
=-1.
lim →1
(x -1) 2
(12) lim
(x -1)
2
x →1
x 2
-x +1=
x lim (x 2
x →1
-x +1)
=0
∴lim
x 2
-x +1x →1
(x -1)
2
=∞.
(13)
log a (1+x )
1
x
=log a (1+x ) x
1
而lim (1+x ) x =e . 而lim log 1x →0
u →e
a u =log a e =
ln a
∴lim
log a (1+x )
=
1x →0
x
ln a
.
(14)令u =a x
-1, 则x =log a (1+u ), 当x →0时，u →0.
x
所以lim
a -1x
==1x →0
lim
u u →0
log a (1+u )
lim
log =ln a (利用(13)题的结果).
a (1+u )
u →0
u
3
3
6x
(15)lim x →0
(1+2x ) sin x =lim x →0
e sin x
ln(1+2x )
=lim e 2x sin x
ln(1+2x )
x →0
1
x 1
=e lim 6⋅
x ⋅ln(1+2x ) 2x
⋅lim ln (1+2
x ) 2x
x →0sin x
=e
6⋅lim
x →0sin
x
x →0
=e
6⨯1⨯ln e
=e 6
.
(16)令u =
sin x sin x x
， 则lim u =lim
x →0
x →0
x
=1
而lim ln u =0 所以lim ln
u →1
x →0
1
sin x x
=0.
27. 利用重要极限lim (1+u ) u =e ，求下列极限：
u →0
x
1⎫⎛⎛x +3⎫
(1)lim 1+⎪; (2)lim ⎪x →∞x →∞
x ⎭⎝⎝x -2⎭
2
2x +1
;
3
2
(3)lim (1+3tan x )
x →0
2cot x
2
; (4)lim (cos2x ) x ;
x →0
(5)lim x [ln(2+x ) -ln x ]; (6)lim
x →∞
1-x ln x
x
x →1
.
1
1
x
1x
22⎡⎛⎡1⎫2⎛1⎫⎤1⎫⎤⎛解：(1)lim 1+⎪=lim ⎢ 1+⎪⎥=⎢lim 1+⎪⎥=e 2=
x →∞x →∞
x ⎭⎝x ⎭⎦x ⎭⎦⎣⎝⎣x →∞⎝
10
5
⎛x +3⎫
(2)lim ⎪x →∞
⎝x -2⎭
2x +1
5⎫⎛
=lim 1+⎪x →∞
x -2⎭⎝
2x +1
x -2
⎡⎤55⎛⎫=lim ⎢1+⎥⎪x →∞ ⎢⎝⎥x -2⎭⎣⎦10
5
5⎫⎛
⋅ 1+⎪
x -2⎭⎝
x -2
⎡⎤55⎛⎫=⎢lim 1+⎥ ⎪⎢⎥x -2⎭⎣x →∞⎝⎦
⎡5⎫⎤⎛10510
⋅⎢lim 1+=e ⋅1=e . ⎪⎥
x -2⎭⎦⎣x →∞⎝
1
3
1
3
(3)lim (1+3tan x )
x →0
2cot x
2
⎤3⎤=⎡2223tan x =lim ⎡=e . 23tan x lim (1+3tan x ) ⎢⎥⎥x →0⎢⎣(1+3tan x ) ⎦⎣x →0⎦
3
1⎧⎫⎪⎪
ln ⎨cos 2x -1⎬2x ⎪[1+(cos2x -1) ]⎪⎩⎭
3
(4)lim (cos2x )
x →0
x
2
=lim e
x →0
x
2
ln cos 2x
3
cos 2x -1
=lim e
x →0
1
3(cos2x -1)
=lim e
x →03lim
x
2
ln [1+(cos2x -1) ]
cos 2x -1
1
cos 2x -1
x
2
=e =e =e
x →0
⋅lim ln [1+(cos2x -1) ]
x →0
cos 2x -1
⎫⎪
⎬⎪⎭
3lim
x →0
-2sin x
x
2
2
1⎧⎪cos 2x -1⋅ln ⎨
lim [1+(cos2x -1) ]⎪x →0⎩
sin x ⎫⎛
-6⋅ lim ⎪⋅ln e
⎝x →0x ⎭
2
=e
-6⨯1⨯1
2
=e .
x
-6
(5)lim x [ln(2+x ) -ln x ]=lim 2⋅
x →∞
x →∞
x 2
⋅ln
2+x x
x
2⎫2⎛
=lim 2ln 1+⎪x →∞
x ⎭⎝
x
⎛⎫2⎫2⎛22⎛⎫=2lim ln 1+⎪=2⋅ln lim 1+⎪
⎪ ⎪x →∞
x ⎭⎝x ⎭⎭⎝x →∞⎝
=2ln e =2.
(6)令x =1+t ，则当x →1时，t →0.
lim
1-x =-1
x →1
ln x
lim
t t →0
ln(1+t )
=-
1
=-
1
1
=-
1ln(1+t ) t
ln ⎡ln e
=-1.
lim t →0
⎢lim (1+t ) t
⎤⎣t →0⎥⎦
28. 利用取对数的方法求下列幂指函数的极限：
1
1x x x
(1)lim (x
x
⎛a +b +c ⎫x
x →0e +x ); (2)lim x →0
⎝3⎪; ⎭
x x
(3)lim ⎛x →∞ ⎝sin 1x +cos 1⎫⎛
1⎫x ⎪⎭; (4)lim x →∞
⎝
1+x 2⎪⎭.
1
解：(1)令y =(ex +x ) x ，则ln y =1x
x
ln(e+x )
于是：
ln x
⎛x ⎫lim x →0ln y =lim 11e +ln 1+x ⎛
x ⎫⎝e x ⎪
⎭ x →0x ln (e x +x )=lim x →0x ln e ⎝
1+e x ⎪⎭=lim x →0x e
x
x
=lim ⎡e 1x
x →0⎢1+⋅⎛
⎣x e ⎝1+x ⎫⎤1⎛x ⎫x ln e x ⎪⎭⎥=1+lim ⎦x →0e x ⋅lim x →0
ln ⎝1+e x ⎪⎭
=1+1⋅ln e =2
即ln (lim y )2
1
=2 x
x →0 即lim x →0
y =e 即lim x →0(e +x )x =e 2. 1
⎛a x +b x x
x x x
(2)令y =+c ⎫x
1a +b +c ，则ln y ⎝3⎪
=⎭
x ln 3于是
x lim (lny 1b x +c
x
x →0
) =lim
x →0
x ln
a +3
a x +b x +c x
-3
1
⎡3
3
=lim ⎛a x +x x ⎫a x +b x +c x -3⎤x →0x ln ⎢⎢⎣ ⎝1+b +c -33⎪⎥⎭⎥⎦
3
x x x
x x x
x
=lim
a +b +c -3
⋅lim ⎛a +b +c -3⎫a
+b x +c x
-3
x →0
3x
x →0
ln ⎝1+3⎪⎭
=1⎛a x
-x
x
⎡3
x x x x x x ⎤3lim x →0 1⎝
x +b -1x +c -1⎫x ⎪⋅ln ⎭⎢⎢lim ⎛⎣x →0 ⎝1+a +b +c -3⎫a +b +c -33⎪⎥⎭⎥⎦
=
13
(lna +ln b +ln c ) ⋅ln e =ln
即lim
x →0
(lny ) =ln
即ln (
lim x →0
y )
=ln
故lim x →0
y =
1
x x
x 即
l ⎛a +b +c ⎫
x
x →i m 0
⎪=.
⎝3⎭
(3)令y =⎛ sin 11⎫x
⎛
11⎫
⎝x +cos x ⎪，则⎭ln y =x ln ⎝
sin x +cos
x ⎪ ⎭
于是
⎛
⎧ 11⎫1
⎝
sin x +cos x -1⎪
⎭lim x →∞ln y =lim x →∞x ln ⎪⎨⎡⎪1+⎛sin 1+cos 1-⎫⎤sin 1+cos 1⎫-1⎪⎩⎢
⎣ ⎝x x 1⎪⎭⎥x x
⎬⎦⎪⎭
1
=lim ⎛11⎫
⎡⎛11⎫⎤sin 1+cos 1x →∞
x ⎝sin x +cos x -1⎪⎭⋅ln ⎢1+⎣
⎝sin x x
-1
x +cos x -1⎪⎭⎥
⎦⎛11⎫⎧=lim sin 1-cos ⎪1
x →∞ -⎪⎪⋅ln ⎨ 11⎪lim ⎡1+⎛⎩⎣ sin 1+cos 1-1⎫⎪⎤sin 1⎫x +cos 1x
-1⎪⎬x →∞⎢⎝x x ⎭⎥⎝x x ⎪
⎭
⎦⎪⎭
⎛ 1⎛1⎫2
⎫sin 1
= lim 2 ⎝x ⎪⎭⎪
⎪⋅ln e =(1-0) ⋅ln e =1 x →∞-lim x →∞ 11⎪
⎝x x ⎪
⎭
即lim ln y =1 从而y x →∞
ln (lim x →∞
)=1 故lim x →∞
y =e x
即 lim ⎛
sin 1+cos 1⎫⎪=e .
x →∞
⎝
x x ⎭x
(4)令y =⎛ ⎝1+1⎫⎛
1⎫x 2⎪，则⎭ln y =x ln ⎝
1+x 2⎪
⎭于是：
1
2
2
lim ⎛
1⎫⎡⎛1⎫x ⎤
x x →∞(lny ) =lim x →∞
x ln ⎝
1+x 2⎪⎭=lim x →∞x ln ⎢⎣ ⎝
1+x 2⎪⎭⎥
⎦x
2
x
2
=lim
1
x →∞
x ln ⎛
1⎫⎝1+x 2⎪⎭
=lim
1
x ⋅lim ln ⎛
11⎫
x →∞
x →∞⎝+x 2⎪⎭
=0⋅ln e =0
即 l i m (l y n =)
0, (l i x →l m
∞
n y x →∞
)= 0
1⎫⎛
x
∴lim →∞y =1 即lim x →∞
⎝
1+x 2⎪=1. x ⎭29. 当x →0时，2x -x 2与x 2-x 3相比，哪个是高阶无穷小量？
x 2
-x 32
解： lim
x →0
2x -x
2
=lim
x -x
x →0
2-x
=0
∴当x →0时，x 2-x 3是比2x -x 2高阶的无穷小量.
30. 当x →1时，无穷小量1-x 与(1)1-x 2, (2)
12
(1-x 2
) 是否同阶？是否等价？解：(1) lim
1-x 1-x
2
=lim
1=
1
x →1
x →1
1+x
2
∴当x →1时，1-x 是与1-x 2同阶的无穷小.
1
(1-x 2
) (2) lim 1+x
x →1
1-x
=lim
x →1
2
=1∴当x →1时，1-x 是与
1
2
(1-x 2
) 等价的无穷小.
31. 利用lim
sin x x
=1或等价无穷小量求下列极限：
x →0
(1)lim
sin m x x →0
sin nx
; (2)lim x →0
x cot x ;
(3)lim
1-cos 2x x 2
x sin x
x →0
; (4)lim
x →0
(5)lim
arctan 3x
n
x x →0
x
; (6)lim n →∞
2sin
2
n
;
2
2
(7)lim
4x -1x →
1arcsin(1-2x ) ; (8)lim
arctan x x →0
;
2
sin
x 2
arcsin x
(9)lim
tan x -sin x -cos βx
x →0
sin x 3
; (10)lim
cos αx x →0
x
2
;
(11)lim
x →0
ln(1-x ) (12)lim
1-cos 4x x →0
2sin 2
x +x t an 2
x
;
ln cos ax ln(sin2
x +e x
(13)lim
) -x x →0
ln cos bx
; (14)lim
x →0
ln(x 2
+e
2x
) -2x
.
解：(1)因为当x →0时，sin mx ~mx , sin nx ~nx ,
所以lim
sin m x sin nx
x →0
=lim
m x nx
x →0
=
m n
.
x cos x lim cos x (2)lim x →0
x cot x =lim
x →0
sin x
⋅cos x =lim
→0
sin x =
x →0
x =1.
x lim sin x x →0
x
2
(3)lim
1-cos 2x x sin x x →0
x sin x
=lim
2sin x →0
x sin x
=2lim
x →0
x
=2.
(4)因为当x →
0时，ln(1+e x sin 2x ) ~e x sin 2x 1~
12
2
x , 所以
x
2
x 2
2
lim
x
x →0
=lim
e sin x 1=lim x →02e ⋅lim ⎛x →0
sin x ⎫
x →0
⎝x ⎪⎭
=2. 2x
2
(5)因为当x →0时，arctan 3x ~3x , 所以
lim
arctan 3x
x
=lim
3x 0
x =3.
x →0x →sin
x (6)lim n
x n
sin
x n
n →∞
2sin
2
n
=lim n →∞
x ⋅
x =x lim
n →∞
x =x .
2
n 2
n (7)因为当x →
12
时，arcsin(1-2x ) ~1-2x , 所以
2
lim
4x -14x 2
-11arcsin(1-2x )
=lim
x →
x →
11-2x
=lim
(2x -1)(2x +1)
1=-lim (2x +1) =-2. 2
2
x →
2
1-2x
x →
12
(8)因为当x →0时，arctan x 2~x 2
, sin
x 2~x 2
, arcsin x ~x , 所以
2
2
lim
arctan x ==.
x →0
lim
x sin
x 0
22
arcsin x
x →x 2
⋅x
(9)因为当x →0时，sin x ~x ,1-cos x ~12
x 2, sin x 3~x 3
, 所以
x ⋅
1
lim
tan x -sin x x
2
x →0
sin x
3
=lim
sin x (1-cos x ) x →0
sin x 3
cos x =lim
x →0
x 3
⋅cos x
=lim
1x →0
2cos x
=12.
(10)因为当x →0时，sin
α+β
α+β
2
x ~
2
x , sin
α-β
2
x ~
α-β
2
x ，所以
α+βlim
cos αx -cos βx
-2sin x sin
α-βx
x →0
x
2
=lim
x →0
x
2
-2⋅α+βα-β=lim 2x ⋅
2
x
x
2
x →=
12
2
2
(β-α).
(11)因为当x →
0时，arcsin ~ln(1-x ) ~-x , 所以
lim
x →0
ln(1-x )
=lim
x →0
-x
=-lim
x →0
=-1. (12)因为当x →0时，sin x ~x , sin 2x ~2x , 所以
2
lim
1-cos 4x 2x x →0
2sin 2
x +x tan 2
x
=lim
2sin x →0
sin 2
x (2+x sec 2
x )
=lim 2⋅(2x )
2x →0
x 2
(2+x sec 2
x )
=lim
82+x sec x
x →0
2
=
8
lim (2+x sec 2
x )
=4.
x →0
(13)因为ln cos ax =ln[1+(cosax -1)],ln cos bx =ln[1+(cosbx -1)], 而当x →0时，cos ax -1→0, cos bx -1→0 故 l n [+1
(c a o x s -
1) ]~a c x -o s +1, l n [b 1x -(c o s b 1x -
) ]又当x →0进，1-cos ax ~1a 2
x 2
,1-cos bx ~
12
2
2
2
b x , 所以
1
ln cos ax
cos ax -1
1-cos ax
a 2x
2
2
lim x →0ln cos bx =lim x →0cos bx -1=lim x →01-cos bx =lim x →01=a
2b 2. 2
b x 2
2
2
(14)因为当x →0时，
sin x e
x
→0,
x e
2x
→0
2
故 ln ⎛ sin 2x ⎫2
~sin x , ln ⎛x 2
⎫
~x , ⎝1+e x ⎪⎭e x ⎝
1+e 2x ⎪⎭e 2x 所以
2x
2x x
ln ⎛ lim
ln(sinx +e ) -x +e ) -ln e ⎝1+sin 2
x ⎫x →0
ln(x 2
+e
2x
) -2x
=lim
ln(sinx x →0
ln(x 2
+e 2x
) -ln e
2x
=lim
e x
⎪⎭x →0
ln ⎛ x 2
⎫⎝
1+
e 2x ⎪⎭sin 2x
x
2
2
=lim
e
x →0
=lim x →0e x ⋅⎛ sin x ⎫x 2⎝⎪=lim x
x →0
e ⋅⎛ ⎝
lim sin x ⎫x ⎭
x →0x ⎪ ⎭e
2x
=e 0
⋅1=1.
32. 求下列函数在指定点处的左、右极限，并说明在该点处函数的极限是否存在？⎧x
(1)f (x ) =⎪⎨x
,
x ≠0, 在x =0处;
⎪⎩1x =0, ⎧x +2, x ≤2(2)f (x ) =⎪
⎨1
=2⎪
⎩x -2
x >0 在x 处.
解：(1)lim f (x ) =lim
x x -x →0
+
x →0
+
x
=lim =1, l i m f x (=) l x
=x →0
+
x
x →0
-
x →
0-
x
→x
-
x
=- 0x
因为 lim f (x ) ≠lim f (x )
x →0
+
x →0
-
所以lim f (x ) 不存在.
x →0
(2)lim f (x ) =lim
1=+∞, lim x →2
+
x →2
+
x -2
-f (x ) =lim -(x +2) =4
x →2
x →2
因为lim +
f (x ) 不存在，所以lim f (x ) 不存在.
x →2
x →2
33. 研究下列函数的连续性，并画出图形：
(1)f (x ) =⎧⎨
x 2,
0≤x ≤1,
(2)f (x ) =⎧x ,
x ≤1, ⎩2-x ,
1
⎩1, x >1;
x
-x =lim
n -n 1-x 2n (3)f (x ) n →∞
n x
+n
-x
; (4)f (x ) =lim
n →∞
1+x
2n
x .
解：(1)由初等函数的连续性知，f (x ) 在（0，1），（1，2）内连续， 又 lim 2
x →1
+f (x ) =lim x →1
+(2-x ) =1, lim x →1
-f (x ) =lim x →1
-x =1
∴lim x →1
f (x ) =1, 而f (1)=1，∴f (x ) 在x =1处连续，
又，由lim 2
+f (x ) =lim +x =0=f (0)，知f (x ) 在x =0处右连续，
x →0
x →0
1
综上所述，函数f (x ) 在[0，2) 内连续. 函数图形如下：
图1-2
(2) 由初等函数的连续性知f (x ) 在(-∞, -1), (-1,1), (1,+∞) 内连续，又由
lim -f (x ) =lim -1=1, lim +f (x ) =lim +x =-1,
x →-1
x →-1
x →-1
x →-1
知lim f (x ) 不存在，于是f (x ) 在x =-1处不连续.
x →-1
-
又由lim f (x ) =lim x =1, lim f (x ) =lim 1=1,
x →1
-
x →1
-
x →1
+
x →1
+
及f (1)=1知lim f (x ) =f (1)，从而f (x ) 在x =1处连续，
x →1
综上所述，函数f (x ) 在(-∞, -1) 及(-1, +∞) 内连续，在x =-1处间断. 函数图形如下：
图1-3
(3)∵当x
n -n n +n
00x x
-x -x
n →∞
=lim
n n
2x 2x
-1+1
n →∞
=-1,
当x =0时，f (x ) =lim
n -n n +n
x 0
n →∞
=0,
当x >0时，f (x ) =lim
n -n n +n
x
-x -x
n →∞
=lim
n n
2x 2x
-1+1
1-=lim
n →∞
11n
2x
n →∞
=1
1+
2x
∴f (x ) =lim
n -n n +n
x
x -x -x
n →∞
⎧-1, ⎪=⎨0, ⎪1, ⎩
x 0.
由初等函数的连续性知f (x ) 在(-∞, 0), (0,+∞) 内连续，
又由 l i m f
x →0
+
x (=)
x →0
l +i m =1 1,
x →0
-
(f l i x m =(-) -l i =m -
x →0
1) 1
知lim f (x ) 不存在，从而f (x ) 在x =0处间断. 综上所述，函数f (x ) 在(-∞, 0), (0,+∞) 内
x →0
连续，在x =0处间断. 图形如下：
图1-4
2n (4)当|x |=1时，f (x ) =lim
1-x
n →∞
1+x 2n
x =0, 1-x 2n 当|x |
1+x
n →∞
2n
x =x , ⎛n
2n 1⎫
当|x |>1时，f (x ) =lim
1-x ⎝x 2⎪⎭-1n →∞
1+x
2n
x =lim
n →∞
⎛n
⋅x =-x
1⎫
⎝x 2⎪⎭
+1⎧x , x
⎨0,
x =1, ⎪⎩
-x , x >1.
由初等函数的连续性知f (x ) 在(－∞, －1), (－1,1),(1,+∞) 内均连续，又由
lim x →-1
-f (x ) =lim -(-x ) =1, lim x →-1
x →-1
+f (x ) =lim x →-1
+x =-1
知lim f (x ) 不存在，从而f (x ) 在x =-1处不连续.
x →-1
又由 l i m f
x (=)
l i -m x (=) - 1,
f l i x m =(1
x →1
+
x →1
+x →1
-
-) x = l i m
x →1
知lim f (x ) 不存在，从而f (x ) 在x =1处不连续.
x →1
综上所述，f (x ) 在(－∞, －1), (－1,1),(1,+∞) 内连续，在x =±1处间断. 图形如下：
图1-5
34. 下列函数在指定点处间断，说明它们属于哪一类间断点，如果是可去间断点，则补充或改变函数的定义，使它连续：
(1)y =(2)y =
x -1x -3x +2x tan x
1x
2
2
2
, x =1, x =2;
π2
, k =0, ±1, ±2, ;
, x =k π, x =k π+, x =0; x ≤1,
(3)y =cos
⎧x -1,
(4)y =⎨
⎩3-x ,
2
x =1. x >1,
解：(1) lim
x -1x -3x +2
2
x →1
=lim
(x -1)(x +1) (x -1)(x -2)
x →1
=-2
lim
x -1x -3x +2
2
2
x →2
=∞
∴x =1是函数的可去间断点. 因为函数在x =1处无定义，若补充定义f (1)=-2，则函
数在x =1处连续；x =2是无穷间断点.
(2) lim
x tan x
=1, lim
x →k π+
x
π2
x →0
tan x
=0
当k ≠0时，lim
∴x =0, x =k π+
π2
x tan x
x →k π
=∞.
π2) =0，
, k =0, ±1, ±2, 为可去间断点，分别补充定义f (0)=1,f (k π+
π2
+k π处连续.(k =0, ±1, ±2, );
可使函数在x =0,及x =
x =k π, k ≠0, k =±1, ±2, 为无穷间断点
(3)∵当x →0时，cos
1x
2
呈振荡无极限，
∴x =0是函数的振荡间断点.(第二类间断点).
(4) lim y =lim (3-x ) =2.
x →1
+
x →1
+
x →1
lim -y =lim -(x -1) =0
x →1
∴x =1是函数的跳跃间断点.(第一类间断点.)
35. 当x =0时，下列函数无定义，试定义f (0)的值，使其在x =0处连续
:
(1)f (x ) =
1x
(2)f (x ) =
tan 2x x
;
1
(3)f (x ) =sin x sin ; (4)f (x ) =(1+x ) x .
解：(1) lim f (x ) =lim
x →0
x →0=lim
x →0
=
32
∴补充定义f (0)=
(2) lim f (x ) =lim
x →0
32
, 可使函数在x =0处连续.
=lim
2x x =2.
tan 2x x
x →0x →0
∴补充定义f (0)=2, 可使函数在x =0处连续.
(3) lim sin x sin
x →0
1x
=0
∴补充定义f (0)=0, 可使函数在x =0处连续.
1
(4) lim f (x ) =lim (1+x ) x =e
x →0
x →0
∴补充定义f (0)=e, 可使函数在x =0处连续.
36. 怎样选取a , b 的值，使f (x ) 在(－∞,+∞) 上连续？
⎧e x ,
(1)f (x ) =⎨
⎩a +x ,
⎧
ax +1, ⎪x
(2)f (x ) =⎨x ≥0; ⎪sin x +b ,
⎪⎩
+
+
x
π2π2
,
.
解：(1) f (x ) 在(-∞, 0), (0,+∞) 上显然连续，而lim f (x ) =lim (a +x ) =a ,
x →0
x →0
x →0
lim -f (x ) =lim -e =1, 且f (0)=a ,
x →0
x
∴当f -(0)=f +(0)=f (0)，即a =1时，f (x ) 在x =0处连续，所以，当a =1时，f (x ) 在
(-∞, +∞) 上连续.
(2) f (x ) 在(-∞, ), (
2
ππ2
, +∞) 内显然连续. 而
lim +f (x ) =lim +(sinx +b ) =1+b ,
x →
π2
x →
π2
lim -f (x ) =lim -(ax +1) =
x →
π2
x →
π2
π2
a +1,
π
f () =1+b , 2
∴当1+b =
π2
a +1，即b =
π2
a 时，f (x ) 在x =
π2
处连续，因而f (x ) 在(-∞, +∞) 上连续.
37. 试证：方程x ⋅2x =1至少有一个小于1的正根.
证：令f (x ) =x ⋅2x -1，则f (x ) 在[0,1]上连续，且f (0)=-10, 由零点定理，
∃ξ∈(0,1)使f (ξ) =0即ξ⋅2-1=0
ξ
即方程x ⋅2x =1有一个小于1的正根.
38. 试证：方程x =a sin x +b 至少有一个不超过a +b 的正根，其中a >0, b >0. 证：令f (x ) =x -a sin x -b ，则f (x ) 在[0,a +b ]上连续， 且 f (0) =-b 0，则由零点定理得.
∃ξ∈(0,a +b ) , 使f (ξ) =0即ξ-a sin ξ-b =0即ξ=a sin ξ+b ，即ξ是方程x =a sin x +b 的根，综上所述，方程x =a sin x +b 至少有一个不超过a +b 的正根.
, ) s i x n ≥
39. 设f (x ) 在[0,2a ]上连续，且f (0)=f (2a ) , 证明：方程f (x ) =f (x +a ) 在[0，a ]内至少有一根.
证：令F (x ) =f (x ) -f (x +a ) , 由f (x ) 在[0,2a ]上连续知，F (x ) 在[0,a ]上连续，且
F (0)=f (0)-f (a ),
F (a ) =f (a ) -f (2a ) =f (a ) -f (0)
若f (0)=f (a ) =f (2a ), 则x =0, x =a 都是方程f (x ) =f (x +a ) 的根，
若f (0)≠f (a ) ，则F (0)F (a )
即f (ξ) =f (ξ+a ) ，即ξ是方程f (x ) =f (x +a ) 的根，
综上所述，方程f (x ) =f (x +a ) 在[0,a ]内至少有一根.
40. 设f (x ) 在[0,1]上连续，且0≤f (x ) ≤1，证明：至少存在一点ξ∈[0,1]，使f (ξ) =ξ. 证：令F (x ) =f (x ) -x , 则F (x ) 在[0,1]上连续，且F (0)=f (0)≥0, F (1)=f (1)-1≤0, 若f (0)=0，则ξ=0, 若f (1)=1, 则ξ=1, 若f (0)>0, f (1)
综上所述，至少存在一点ξ∈[0,1]，使f (ξ) =ξ.
41. 若f (x ) 在[a , b ]上连续，a
f (ξ) =
f (x 1) +f (x 2) + +f (x n )
n
.
证: 由题设知f (x ) 在[x 1, x n ]上连续，则f (x ) 在[x 1, x n ]上有最大值M 和最小值m ，于是
m ≤
f (x 1) +f (x 2) + +f (x n )
n
≤M ,
由介值定理知，必有ξ∈[x 1, x n ]，使
f (ξ) =
f (x 1) +f (x 2) + +f (x n )
n
.